User:Raul654/proof

Proof that 0 = -1

I stumbled across this in my high school Calculus BC class. Here's a challenge for my fellow math geeks - what is wrong here?

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Given:

$\int _{}u\,dv=u*v-\int _{}v\,du$

Let $\mathbf {dv} =\mathbf {sin(x)*dx}$

Therefore, $\mathbf {v} =-\mathbf {cos(x)}$

Let $\mathbf {u} =\mathbf {sec(x)}$

Therefore, $\mathbf {du} =\mathbf {sec(x)*tan(x)*dx}$

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$\int _{}\tan(x)\,dx=\int _{}\tan(x)\,dx$

$\int _{}\tan(x)\,dx=\int _{}\sec(x)*\sin(x)\,dx$

Substitute from above.

$\int _{}\tan(x)\,dx=\sec(x)*(-\cos(x))-\int _{}-\cos(x)*\sec(x)*\tan(x)\,dx$

Sec(x) * cos(x) = 1

$\int _{}\tan(x)\,dx=-1-\int _{}-\tan(x)\,dx$

Group the minus signs.

$\int _{}\tan(x)\,dx=-1+\int _{}\tan(x)\,dx$

(Subtract the integral from each side)

$\mathbf {0} =\mathbf {-1}$

Voilà - a logical inconsistency.